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2 years ago

Luna simplifies the expression (2x-4)-(3x+2) correctly. which expression represents luna's result

1 answer:

5 0

Okay, so first you have to factor out that negative throughout the parenthesis.
(2x-4)-3x-2
Then, put like terms together so you can organize your problem.
2x-3x-4-2
Now all you have to do is combine the like terms!
-x-6
Oh look, that's a. A is the answer.

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A building 220 feet casts a 100 foot long shadow. If a person stands at the end of the shadow and looks up to the top of the bui

kaheart [24]

Answer:

65.56°

Step-by-step explanation:

= We solve the above question using the Trigonometric function of Tangent

tan θ = Opposite/Adjacent

Opposite = 220 feet = Height of the building

Adjacent = 100 feet = Length of the shadow

Hence,

tan θ = 220/100

tan θ = 2.2

= arctan(2.2)

= 65.55604522°

Approximately = 65.56°

5 0

2 years ago

You have a die with two red faces, two blue faces, and two white faces. What is the probability the die is tossed and shows eith

elena-s [515]

Answer:

1/3??? im probably wrong but i would answer it like this.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

HELP PLEASE! a + 2 2/5 = 4 3/10

amm1812

Answer:

Step-by-step explanation:

4 + 0 = 0

7 0

2 years ago

FR.EE POINTS HAVE A GOOD DAY 100PTS :)

Law Incorporation [45]

Answer:

Thank u so much have a good day too

5 0

2 years ago

Read 2 more answers

A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) Two computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) Eight computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

LET X = Number of computers classified as cutting dash edge

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

7 0

2 years ago